Find the global maxima/minima of $f(x,y,z) = x+y+z$ for points inside of $A = \{ (x,y,z) \in \mathbb{R}^3: x^2-y^2 = 1 \wedge 2x+z = 1 \}$
I renamed the conditions of $A$ to a function $g(x,y,z) = x^2-y^2-2x-z = 0$ in order to be able to use Lagrange multipliers.
Deriving $f$:
$\nabla(x,y,z) = (1,1,1)$
Deriving $g$:
$g_x = 2x-2$
$g_y = -2y$
$g_z = -1$
By solving the Lagrange system:
\begin{cases} 1 = \lambda(2x-2) \\ 1 = \lambda (-2y) \\ 1 = \lambda (-1) \\ g(x,y) = 0 \end{cases}
I get that $(x,y,z) = (\frac{1}{2}, \frac{1}{2}, -1)$ for $\lambda = -1$. But that point is not inside A.
Is this enough to guarantee that $f|_A$ does not have either maxima nor minima?
(Since the second derivatives are all zero, the second derivative test doesn’t give any information.)
Thanks!
EDIT:
By parametrizing $y = \pm \sqrt{x^2-1}$ and $z=1-2x$ and doing the composition with $f$ I get that $f(x, \sqrt{x^2-1}, 1-2x)$ is monotonically increasing and $f(x, -\sqrt{x^2-1}, 1-2x)$ monotonically decreasing. Would that prove that $f$ never reaches either maxima nor minima?
$$
f(-t,\sqrt{t^2-1},1+2t)=1+t+\sqrt{t^2-1}
$$
which goes to $+\infty$ as $t\to\infty$
$$
f(t,-\sqrt{t^2-1},1-2t)=1-t-\sqrt{t^2-1}
$$
which goes to $-\infty$ as $t\to\infty$
So there appears to be no maximum or minimum.
You are correct in observing that it is a warning sign when the “Lagrange equations” cannot be solved consistently. The values $ \ ( x, \ y ) \ = \ ( \frac{1}{2} , \ \frac{1}{2} ) \ $ do not lie on the constraint hyperbola, nor does $ \ (x, \ z) \ = \ (\frac{1}{2}, \ -1) \ $ sit in the constraint plane. Clearly something is amiss.
Because the Lagrange-multiplier method locates points where the directions of the normal vectors to level “curves” or “surfaces” of the function $ \ f \ $ and the constraint “curve” or “surface”, the calculations will be inconsistent if there are no such points.
If we consider a geometrical interpretation for this problem, we can see another way of understanding the difficulty. The plane $ \ 2x \ + \ z \ = \ 1 \ $ intersects the hyperbolic cylinder $ \ x^2 \ – \ y^2 \ = \ 1 \ $ . If we apply the two constraints to describe the intersection curve in $ \ \mathbb{R}^3 \ $ , we find
$$ \left[ \ \frac{1}{2} \ (1 – z) \ \right]^2 \ – \ y^2 \ = \ 1 \ \ \Rightarrow \ \ ( z – 1 )^2 \ – \ 4y^2 \ = \ 4 \ \ . $$
The projections of this curve onto either the $ \ xy- \ $ or the $ \ xz-$ plane are hyperbolas with the center at $ \ (x, \ y,\ z) \ = \ ( 0, \ 0, \ 1) \ $ . Without any other constraint provided on permissible points, such as requiring an extremum to be in the first octant, the opening of the “branches” of the hyperbola away from the origin will not impose any limit in the positive or negative direction for the function $ \ f(x, \ y, \ z) = x+y+z \ $ . There are no solution extrema because, in a sense, there is no “inside” to the region described: the boundary is an open curve.
Here is a graph of the situation.
The same problem was asked later, with the calculation made using two “multipliers”; similar trouble arises there.